BxMO Team Selection Test

Denote the radius of the incircle of $△ABC$ by $r$. Then the area of triangle $ABC$ is $$\frac{1}{2}|AB|\cdot r+\frac{1}{2}|BC|\cdot r+\frac{1}{2}|AC|=\frac{1}{2}r\cdot (3|BC|+|BC|)=2r|BC|.$$ On the other hand, the area of $ABC$ equals $\frac{1}{2}h|BC|$, where $h$ is the altitude from $A$. Hence, $h=4r$. Because the distance from $KL$ to $BC$ is exactly $2r$, the distance from $A$ to $KL$ is also $2r$. Triangles $AKL$ and $ABC$ are similar, because $KL\parallel BC$, and the altitudes from $A$ have lengths $2r$ and $4r$, respectively, giving a multiplication factor of exactly 2. Hence, $K$ is the midpoint $AB$, and $L$ is the midpoint of $AC$. For the point $T$, we have $|AC|=4|AT|$, hence $|AT|=\frac{1}{4}|AC|=\frac{1}{2}|AL|$, hence $T$ is the midpoint of $AL$. Now consider triangle $ABL$. In this triangle, the segment $BT$ is a median, because $T$ is the midpoint of $AL$. Also $LK$ is a median as $K$ is the midpoint $AB$. Their intersection point $S$ is the centroid, from which we get that $\frac{|SL|}{|KL|}=\frac{2}{3}$. $\square$